$A$ metal crystallizes in a simple cubic lattice. The volume of one unit cell is $6.4 \times 10^7 \ pm^3$. What is the radius of the metal atom in $pm$?

An element crystallizes in a body-centred cubic lattice. The edge length of the unit cell is $200 \ pm$ and the density of the element is $5.0 \ g \ cm^{-3}$. Calculate the number of atoms in $100 \ g$ of this element.

The density of chromium metal is $7.2 \, g \, cm^{-3}$. If the edge length of the unit cell is $289 \, pm$, determine the type of unit cell (Simple Cubic, Body-Centered Cubic, or Face-Centered Cubic). [Atomic mass of $Cr = 52 \, a.m.u.$, $N_A = 6.02 \times 10^{23} \, mol^{-1}$]

Calculate the molar mass of an element having density $7.8 \ g \ cm^{-3}$ that forms a $bcc$ unit cell. $\left[a^3 \cdot N_{A} = 16.2 \ cm^3 \ mol^{-1}\right]$

Calculate the volume of a $bcc$ unit cell if the radius of an atom present in it is $1.86 \times 10^{-8} \ cm$.

Calculate the number of unit cells present in $1 \text{ g}$ of a metal if the product of the density of the metal and the volume of the unit cell is $2.5 \times 10^{-22} \text{ g}$.